题目：https://leetcode-cn.com/problems/ji-qi-ren-de-yun-dong-fan-wei-lcof/

代码：

class Solution {
    public int movingCount(int m, int n, int k) {
        LinkedList<Point> queue = new LinkedList<>();
        int[][] visited = new int[m][n];
        queue.offerLast(new Point(0, 0));
        visited[0][0] = 1;
        int count = 0;
        Point p;
        while(!queue.isEmpty()){
            p = queue.pollFirst();
            count ++;
            if(p.x + 1 < m && visited[p.x + 1][p.y] == 0 && check(p.x+1, p.y, k)){
                queue.offerLast(new Point(p.x+1, p.y));
                visited[p.x+1][p.y] = 1;
            }
            if(p.y+1 < n && visited[p.x][p.y + 1] == 0 && check(p.x, p.y+1, k)){
                queue.offerLast(new Point(p.x, p.y+1));
                visited[p.x][p.y+1] = 1;
            }
        }
        return count;
    }
    private boolean check(int x, int y, int k){
        int sum = 0;
        while(x > 0){
            sum += x % 10;
            x /= 10;
        }
        while(y > 0){
            sum += y % 10;
            y /= 10;
        }
        return sum <= k;
    }
    class Point{
        int x, y;
        public Point(int x, int y){ this.x = x; this.y = y;}
    }
}

执行用时：4 ms, 在所有 Java 提交中击败了26.55%的用户

内存消耗：36.2 MB, 在所有 Java 提交中击败了23.34%的用户

优化：上面方法中，机器人每走一步都要判断该步是否符合条件（调用ckeck方法），但实际上每一步都只移动一个方向，这样不需要每次都重新计算纵方向和横方向的各个位数之和。

class Solution {
    public int movingCount(int m, int n, int k) {
        LinkedList<Point> queue = new LinkedList<>();
        int[][] visited = new int[m][n];
        queue.offerLast(new Point(0, 0, 0, 0));
        visited[0][0] = 1;
        int count = 0;
        Point p;
        while(!queue.isEmpty()){
            p = queue.pollFirst();
            count ++;
            if(p.x + 1 < m && visited[p.x + 1][p.y] == 0){
                int t = p.x % 10 == 9 ? p.xSum - 8 : p.xSum + 1;
                if(t + p.ySum <= k){
                    queue.offerLast(new Point(p.x+1, p.y, t, p.ySum));
                    visited[p.x+1][p.y] = 1;
                } 
            }
            if(p.y + 1 < n && visited[p.x][p.y + 1] == 0){
                int t = p.y % 10 == 9 ? p.ySum - 8 : p.ySum + 1;
                if(t + p.xSum <= k){
                    queue.offerLast(new Point(p.x, p.y+1, p.xSum, t));
                    visited[p.x][p.y+1] = 1;
                }
            }
        }
        return count;
    }
    class Point{
        int x, y;
        int xSum, ySum;
        public Point(int x, int y, int xSum, int ySum){ 
            this.x = x; 
            this.y = y;
            this.xSum = xSum;
            this.ySum = ySum;
        }
    }
}

执行用时：3 ms, 在所有 Java 提交中击败了30.20%的用户 内存消耗：36.1 MB, 在所有 Java 提交中击败了26.73%的用户